Question: why are there no non-trivial extensions of $\mathbb{G}_m$ by abelian varieties?

Specifically, let $A$ be an abelian variety over a field $k$. Then it seems to be well known that any extension $$ 0 \to A \to G \to \mathbb{G}_m \to 0$$ in the category of group varieties over $k$ is split. For instance, this is mentioned in Deligne's Theorie de Hodge I page 426 just before Section 3. Chevalley's structure theorem of algebraic groups certainly implies this.

Question: Is there a simple/direct proof?

Note that there are non-trivial extensions of abelian varieties by tori (semiabelian varieties).

I would have thought that Chevalley's theorem on the structure of connected algebraic groups would be an ideal answer to the question. To me, it seems easier than trying to compute an Ext group...
–
Pete L. ClarkMay 10 '11 at 4:07

2

How, does Chevalley's theorem imply this? I can see how it implies that the extension splits after pulling back along some $n$'th power map $\mathbb{G}_m \to \mathbb{G}_m$ but not that it is actually trivial.
–
Torsten EkedahlMay 10 '11 at 4:18

@Pete: Thanks for the comment. @Ekedahl: You are right, I misspoke! Thanks for clarifying!
–
SGPMay 10 '11 at 13:06

1 Answer
1

Using the Kummer exact sequence
$0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we
get a long exact sequence
$$
0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)
\rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A).
$$
As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding
$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at
least over an algebraically closed field in which $n$ is invertible we always
have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently so is
$\mathrm{Ext}^1(\mathbb{G}_m,A)$.

This does not contradict Deligne's claim as he seems to be (somewhat implicitly
it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this
is made more clear). In fact Chevalley's theorem implies that this is true:
Consider a connected affine subgroup $G'\hookrightarrow G$ for which the
quotient is an abelian variety. Then the kernel of the composite
$G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On
the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$
would be finite and hence trivial so $G$ would be an abelian variety which is
not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then
some $\mu_n$ which shows that the extension is in the image of
$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in
particular is trivial up to isogeny.

The final conclusion is that $\mathrm{Ext}^1(\mathbb{G}_m,A)=\mathrm{Hom}(\hat{\mathbb Z}(1),A)$.

Addendum: I just realised that I didn't answer the actual question about
whether there is a simpler proof (to the isogeny statement would be my
interpretation). I doubt it (depending of course somewhat on your definition of
simpler). A proof (arguably less simple) avoiding the use of Chevalley's theorem
(but assuming to be on the safe side that the base field is algebraically closed
of characteristic $0$) would be to prove that for some $n$ the pull back of the
extension along the $n$'th power on $\mathbb{G}_m$ is trivial as an $A$-torsor
because then the extension would be described by a $2$-cocycly
$\mathbb{G}_m\times\mathbb{G}_m\rightarrow A$ and any map of that form is
constant. That in turn will follow from the fact that for some $m$ the class of
this extension as an $A$-torsor would be in the image of
$H^1(\mathbb{G}_m,A[m])\rightarrow H^1(\mathbb{G}_m,A)$. This follows from the
fact that $H^1(\mathbb{G}_m,A)$ is torsion and then the conclusion follows as
any class of $H^1(\mathbb{G}_m,C)$ for any finite locally constant sheaf $C$ is
killed by pulling back along some $n$'th power map.